In Shannon formula below: $$R = W\log_2\left(1 +\frac{P_t\cdot H^2}{W N_0}\right)$$ why square the channel gain? When calculating the channel capacity, why do we need to square the channel gain?

Notation: $W$ is the bandwidth, $P_t$ is the transmit power, $H$ is the channel gain and $N_0$ is the thermal noise power density.

  • $\begingroup$ Without assuming understand all your notations, the channel gain, if it were there, should be squared because the Shannon AWGN capacity equation contains power/energy terms, and power/energy terms are square of absolute gain. $\endgroup$
    – AlexTP
    Commented Jun 10, 2017 at 10:47

1 Answer 1


The Shannon channel capacity equation, as I remember is

$$\begin{align} R &= W \log_2 \left( 1 + \frac{S}{N} \right) \\ \\ &= W \log_2 \left( \frac{S+N}{N} \right) \\ \\ &= W \big( \log_2(S+N) \, - \, \log_2(N) \big) \\ \\ &= \frac{W}{6.02} \big( (S+N)_\text{dB} \, - \, (N)_\text{dB} \big) \\ \end{align}$$

where $S$ is the signal power and $N$ is the noise power (both over the same bandwidth $W$).

Noise power for a given bandwidth (thin enough that noise power density $N_0$ is considered constant is

$$ N = N_0 W $$

and the received signal power is the transmitted power, $P_t$, times the power gain of the channel (assuming both are constant over the bandwidth $W$). The power gain of the channel is the square of the gain in amplitude because instantaneous power is the square of instantaneous amplitude. So the received signal power is

$$ S = |H|^2 P_t $$

assuming all of the transmitted power is in the channel bandwidth $W$ and the channel gain, $H$, is constant over the bandwidth $W$.


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